Answer:
As SI unit of energy, \[J=kg{{m}^{2}}{{s}^{-2}},\]so
[Energy] \[=[M{{L}^{2}}{{T}^{-2}}]\]
(a) \[[{{m}^{2}}{{\upsilon }^{2}}]=[{{M}^{2}}]{{[L{{T}^{-1}}]}^{2}}=[{{M}^{2}}{{L}^{2}}{{T}^{-2}}]\]
(b) \[[1/2m{{\upsilon }^{2}}]=[M]{{[L{{T}^{-1}}]}^{2}}=[M{{L}^{2}}{{T}^{-2}}]\]
(c) \[[ma]=[M][L{{T}^{-2}}]=[ML{{T}^{-2}}]\]
(d) \[[3/16m{{\upsilon }^{2}}]=[M]{{[L{{T}^{-1}}]}^{2}}=[M{{L}^{2}}{{T}^{-2}}]\]
(e) The quantities \[\left( \text{1}/\text{2} \right)m{{\upsilon }^{2}}\]and ma have different dimensions and hence cannot be added.
Since the kinetic energy K has the dimensions of\[[M{{L}^{2}}{{T}^{-2}}],\] formulas (a), (c) and (e) are clearly ruled out.
Dimensional analysis cannot tell which of the two, (b) or (d), is the correct formula. From the actual definition of kinetic energy, only (b) is the correct formula for kinetic energy.
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